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I know that there are at least 3 other questions with the similar problem. I hope, though, you won't flag this as a duplicate since I've got some specific questions here that haven't been answered there before.

So we all know the theorem states:

The sequence of functions $\{f_n\}_{n\in \mathbb{N}}$ converges uniformly to the function $f$ if and only if $$\lim_{n\to \infty}\sup{|f_n(x)-f(x)|} = 0$$

And the proof that I've found says the following:

For the right hand side implication:

Let $l_n = \sup{|f_n(x)-f(x)|}$. From the assumtion (that the function converges uniformly) we know that:

$$(\forall\epsilon>0)(\exists n_0 \in \mathbb{N})(\forall x \in X)(n \geq n_0 \Rightarrow |f_n(x)-f(x)|) < \frac{\epsilon}{2}$$

Therefore $l_n \leq \frac{\epsilon}{2} < \epsilon \Rightarrow \lim_{n \to \infty}l_n = 0$. Proof done.

Here's what confuses me:

The fact that $l_n$ is the supremum of the set with elements $|f_n(x)-f(x)|$ tells us that it is always greater than those elements. How does that tell us that it's less than $\epsilon$. The only justification that I can come up with is to recall that the supremum is the least upper bound, and since $\epsilon$ is also an upper bound in this case, than the supremum must be less than it. But still not sure how valid that is.

Another thing: If my justification is correct, why do we even use $\frac{\epsilon}{2}$? Can't we just immediately conclude that $l_n$ must be less than $\epsilon$. It seems unnecessary.

And finally, the textbook I'm reading this from is missing the proof for the left hand side implication. So could anyone give me some hints on how to prove that side?

Thanks.


EDIT:

Some similar questions I've found: Uniform convergence of supremum Uniform Convergence Theorem

There was one more, I think, but I can't find it anymore.

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  • $\begingroup$ please link the "at three other questions" that you found that are similar to this question. $\endgroup$ – jordan_glen Jan 28 at 15:07
  • $\begingroup$ @jordan_glen done $\endgroup$ – Koy Jan 28 at 15:16
  • $\begingroup$ Thanks, dear @Key! $\endgroup$ – jordan_glen Jan 28 at 15:18
  • $\begingroup$ Congrats, you've posed question number $2^{64}$. Made the MB. $\endgroup$ – Michael Hoppe Jan 28 at 15:25
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You were right about the justification. If we have a set $A$ and we know $a < b$ for all $a \in A$ we know that $b$ is an upperbound for $A$ and so $\sup A \le b$ because $\sup A$ is the minumum of the upperbounds. (a set like $A=(0,1)$, and $b=1$ shows we can have equality here, so $\le$ is the best we can do).

We apply this in the proof to the set $A=\{|f_n(x)-f(x)|: x \in X\}$ ($X$ or whatever the domain of all $f_n$ and $f$ is) and $b=\frac{\varepsilon}{2}$. Then $\sup A=l_n \le \frac{\varepsilon}{2} < 2$. So the half factor is just to get $<$ in the end as we only get $\le$ from the $\sup$ argument.

As to the other direction: suppose $l_n \to 0$ and let $\varepsilon >0$ be given. Then find $n_0$ such that $n \ge n_0$ implies that $l_n < \varepsilon$, by the given limit.

If then $n \ge n_0$ and $p\in X$, then $|f_n(p) -f(p)| \le \sup \{|f_n(x)-f(x)|: x \in X\}=l_n$ (any element is $\le$ than the sup, which is in particular an upper bound of the set) and so $|f_n(p) - f(p)| \le l_n < \varepsilon$, and as $n \ge n_0$ and $p \in X$ were arbitary we have shown uniform convergence of $f_n$ to $f$.

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